(a) Define uniform acceleration. (b) Forces act on a car in motion. List the (i) horizontal forces and their directions; (ii) vertical forces and their dire...
(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\(^{-1}\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.
(i) Sketch a velocity-tune graph for the motion.
(ii) Use the graph to determine the (\(\alpha\)) total distance travelled by the car (\(\beta\)) deceleration of the car.
The figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
The weight of the car, acting vertically downward.
The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
The weight of the car, acting vertically downward.
The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
